Proof problem: show that $n^a < a^n$ for all sufficiently large n

Question

I would like to show that $n^a < a^n$ for all sufficiently large $n$, where $a$ is a finite constant.

This is clearly true by intuition/graphing, but I am looking for a rigorous proof. Can anyone help me out? Thanks.

Answer 1

Set $f(n)=\frac{a^n}{n^a}$; we seek to prove $f(n)>1$ for all $n$ sufficiently large. We rewrite this using one of the laws of exponents as $$f(n)=\frac{e^{n \ln a}}{e^{a\ln n}}=e^{n\ln a-a\ln n}=e^{\ln n\left(\frac{n}{\ln n}\ln a - a\right)}$$

Hence we need only choose $n$ large enough that $\frac{n}{\ln n}\ln a - a>0$, i.e. such that $$\frac{n}{\ln n}>\frac{a}{\ln a}$$ It is easy to prove that $\frac{n}{\ln n}$ is an increasing function for $n\ge e$ (proof on request), hence for all $n>\max(a,e)$ this is true.

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PS: The statement is false if $a\le 1$, so we must assume $a>1$.

Answer 2

Take $\log$ from both sides $a \log n < n \log a $

Now lets check who grows faster $\lim\limits_{n\to \infty}\frac{n \log a}{a \log n}=\infty$ and $\lim\limits_{n\to \infty}\frac{a \log n}{n \log a}=0$

Answer 3

It is enough to prove $\log(n^a)=a\log n<\log(a^n)n\log a$ for all sufficiently large $n$, i.e. $$\frac{\log n}n<\frac{\log a}a,$$ which is true for all sufficiently large $n$ since $\dfrac{\log n}n\to 0$ as $n\to\infty$.

Answer 4

Verify that the inequality fails if $a\le 1.$ So we assume $a>1.$

Let's first show $n < a^n$ for large $n.$ By the binomial theorem, for $n\ge 2,$

$$a^n = (1+(a-1))^n = 1 + n(a-1) + n(n-1)(a-1)^2/2 + \cdots \ge n(n-1)(a-1)^2/2.$$

Clearly the last expression is $>n$ for large $n.$

To show $n^a < a^n$ for large $n,$ note that $a^{1/a}> 1.$ By the above, $n<(a^{1/a})^n$ for large $n.$ This is the same as saying $n<(a^n)^{1/a},$ or $n^a < a^n,$ for large $n.$

Answer 5

What you want is $n^{1/n} < a^{1/a}$.

Since $x^{1/x}$ has a max at $x=e$ and is decreasing for $x > e$, $n^{1/n} < a^{1/a}$ if $e \le a < n$.

If $0 < a < 1$, then, since $a^{1/a} < 1$ and $n^{1/n} > 1$ for $n > 1$, $n^{1/n} > a^{1/a}$ so what you want does not happen.

If $a > 1$, since $a^{1/a} > 1$ and $n^{1/n} \to 1$ as $n \to \infty$, what you want does happen for large enough $n$.

To get a simple bound on $n^{1/n}$:

By Bernoulli's inequality, $(1+n^{-1/2})^n > 1+n(n^{-1/2}) =1+n^{1/2} \gt n^{1/2} $. Raising both sides to the $2/n$ power, $(1+n^{-1/2})^2 \gt n^{1/n} $ or $n^{1/n} < (1+n^{-1/2})^2 < 1+3n^{-1/2} $ for $n > 1$.

Therefore a sufficient condition for $n^{1/n} < a^{1/a}$ is $a^{1/a} > 1+3n^{-1/2} $ or $3n^{-1/2} < a^{1/a}-1 $ or $9/n < (a^{1/a}-1)^2 $ or $n > \dfrac{9}{(1+a^{1/a})^2} $.